Is there a name for a binomial expansion without coefficients?

Question

I am investigating a problem from George E. Andrews Number Theory (Dover, 1971), discussed previously here:

Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$

I was led astray for a bit, because I misunderstood the rhs to be the difference of $x$ and $y$ times the binomial expansion $(x+y)^{n-1}$.

But this is wrong... the rhs is actually the difference of $x$ and $y$ times the binomial expansion $(x+y)^{n-1}$ "without the coeffients", i.e.,

$$\sum_{i=0}^{n-1}x^{n-1-i}y^i$$

(also briefly discussed here: Binomial summation without coefficient)

Is there a name for this summation?

Answer 1

There is no specific name for the bivariate polynomial \begin{align*} x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1} \end{align*}

but the nice expression \begin{align*} (x-&y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\ \end{align*}

is called a telescoping sum since all terms besides the first and the last cancel out.

\begin{align*} (x-&y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\ &=x(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\ &\qquad-y(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\\ &=\color{blue}{x^{n}}+x^{n-1}y+x^{n-2}y^2+\cdots+x^3y^{n-3}+x^2y^{n-2}+xy^{n-1}\\ &\qquad\ \;-x^{n-1}y-x^{n-2}y^2-\cdots-x^3y^{n-3}-x^2y^{n-2}-xy^{n-1}\color{blue}{-y^{n}}\\ &=\color{blue}{x^n-y^n}\\ \end{align*}

Answer 2

I'm not sure if I'm answering your question here, but this is what I usually think:

\begin{align} &(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})\\ \\ &=(x^n+x^{n-1}y+\cdots+x^2y^{n-2}+xy^{n-1})-(x^{n-1}y+x^{n-2}y^2+\cdots+xy^{n-1}+y^n)\\ \\ &=x^n+(x^{n-1}y-x^{n-1}y)+(x^{n-2}y^2-x^{n-2}y^2)+\cdots+(xy^{n-1}-xy^{n-1})-y^n\\ \\ &=x^n-y^n \end{align}

Of course, this uses induction implicitly, but its not hard to see why it works from the standpoint of distributing and everything canceling out.